python中的设施位置问题与纸浆

Question

I do a facility location problem solution with pulp in python, it calculates the solution path correctly, but there is a problem with the total cost.

from pulp import *

Customer = [1, 2, 3, 4]
Facility = ['Fac-1', 'Fac-2', 'Fac-3']
Demand = {1: 50, 2: 50, 3: 75, 4: 75}
Max_Supply = {'Fac-1': 100, 'Fac-2': 100, 'Fac-3': 500}
fixed_cost = {'Fac-1': 100.123, 'Fac-2': 100.456, 'Fac-3': 100.789}
transportation_cost = {'Fac-1': {1: 100.1, 2: 100.4, 3: 200.7, 4: 200.1}, 'Fac-2': {1: 200.2, 2: 200.5, 3: 100.8, 4: 200.11}, 'Fac-3': {1: 2000.3, 2: 2000.6, 3: 2000.9, 4: 100.12}}

prob = LpProblem("Capacitated_Facility_Location_Problem", LpMinimize)

use_facility = LpVariable.dicts("Use Facility", Facility, 0, 1, LpBinary)

ser_customer = LpVariable.dicts("Service", [(i,j) for i in Customer for j in Facility], 0)

prob += lpSum(fixed_cost[j]*use_facility[j] for j in Facility) + lpSum(transportation_cost[j][i]*ser_customer[(i,j)] for j in Facility for i in Customer)

for i in Customer:
    prob += lpSum(ser_customer[(i,j)] for j in Facility) == Demand[i]

for j in Facility:
    prob += lpSum(ser_customer[(i,j)] for i in Customer) <= Max_Supply[j]*use_facility[j]

for i in Customer:
    for j in Facility:
        prob += ser_customer[(i,j)] <= Demand[i]*use_facility[j]

for v in prob.variables():
    if v.varValue > 0:
        print(v.name, "=", v.varValue)

print("Total Cost = ", value(prob.objective))

Output:

• Service_(1,_'Fac_1') = 50.0
• Service_(2,_'Fac_1') = 50.0
• Service_(3,_'Fac_2') = 75.0
• Service_(4,_'Fac_3') = 75.0
• Use_Facility_Fac_1 = 1.0
• Use_Facility_Fac_2 = 1.0
• Use_Facility_Fac_3 = 1.0
• Total Cost = 25395.368000000002

but total cost should be like below

• transport cost = (100.1 + 100.4 + 100.8 + 100.12) = 401.42

• fixed_cost = (100.123 + 100.456 + 100.789) = 301.368
• Total Cost = 702.708

Answer 1

The computer is correct....go figure. :)

In your objective, you are multiplying the transport cost times the service amount. See the last line of the output below. Also note how you can "break up" the expression for ease in troubleshooting and then just use the parts in the objective function (or elsewhere).

from pulp import *

Customer = [1, 2, 3, 4]
Facility = ['Fac-1', 'Fac-2', 'Fac-3']
Demand = {1: 50, 2: 50, 3: 75, 4: 75}
Max_Supply = {'Fac-1': 100, 'Fac-2': 100, 'Fac-3': 500}
fixed_cost = {'Fac-1': 100.123, 'Fac-2': 100.456, 'Fac-3': 100.789}
transportation_cost = {'Fac-1': {1: 100.1, 2: 100.4, 3: 200.7, 4: 200.1}, 'Fac-2': {1: 200.2, 2: 200.5, 3: 100.8, 4: 200.11}, 'Fac-3': {1: 2000.3, 2: 2000.6, 3: 2000.9, 4: 100.12}}

prob = LpProblem("Capacitated_Facility_Location_Problem", LpMinimize)

use_facility = LpVariable.dicts("Use Facility", Facility, 0, 1, LpBinary)

ser_customer = LpVariable.dicts("Service", [(i,j) for i in Customer for j in Facility], 0)

fixed = lpSum(fixed_cost[j]*use_facility[j] for j in Facility)
transpo = lpSum(transportation_cost[j][i]*ser_customer[(i,j)] for j in Facility for i in Customer)  
prob += fixed + transpo

for i in Customer:
    prob += lpSum(ser_customer[(i,j)] for j in Facility) == Demand[i]

for j in Facility:
    prob += lpSum(ser_customer[(i,j)] for i in Customer) <= Max_Supply[j]*use_facility[j]

for i in Customer:
    for j in Facility:
        prob += ser_customer[(i,j)] <= Demand[i]*use_facility[j]

prob.solve()
for v in prob.variables():
    if v.varValue > 0:
        print(v.name, "=", v.varValue)

print("Total Cost = ", value(prob.objective))
print(f'fixed: {value(fixed)}')
print(f'transpo: {value(transpo)}')
t = transportation_cost["Fac-1"][1] * ser_customer[1, "Fac-1"].varValue
print(f'the transpo for fac 1 to cust 1 is: {t}')

Output:

Service_(1,_'Fac_1') = 50.0
Service_(2,_'Fac_1') = 50.0
Service_(3,_'Fac_2') = 75.0
Service_(4,_'Fac_3') = 75.0
Use_Facility_Fac_1 = 1.0
Use_Facility_Fac_2 = 1.0
Use_Facility_Fac_3 = 1.0
Total Cost =  25395.368000000002
fixed: 301.368
transpo: 25094.0
the transpo for fac 1 to cust 1 is: 5005.0